# Primitive Roots of Unity

**Primitive $n^\text{th}$ roots of unity** are roots of unity whose multiplicative order is $n.$ They are the roots of the $n^\text{th}$ cyclotomic polynomial, and are central in many branches of number theory, especially algebraic number theory.

## Definition

Let $n$ be a positive integer. A

primitive $\ n^\text{th}$ root of unityis an $n^\text{th}$ root of unity that is not a $k^\text{th}$ root of unity for any positive $k<n.$ That is, $\zeta$ is a primitive $n^\text{th}$ root of unity if and only if$\zeta^n=1, \text{and } \zeta^k \ne 1 \text{ for any positive integer } k < n.$

There are four $4^\text{th}$ roots of unity given by $\pm 1, \pm i.$ Two of these, namely $\pm i,$ are primitive. The other two are not: $1^1 = 1$ and $(-1)^2=1.$

## Basic Properties

The primitive $n^\text{th}$ roots of unity are the complex numbers

$e^{2\pi i k/n} : 1\le k \le n, \text{gcd}(k,n)=1.$

There are $\phi(n)$ primitive $n^\text{th}$ roots of unity, where $\phi(n)$ is Euler's totient function.

Let $\zeta_n = e^{2\pi i/n}.$ Recall that the $n^\text{th}$ roots of unity are the $n$ distinct powers $\zeta_n^k = e^{2\pi i k/n} : 1 \le k \le n.$ So it remains to show that $\zeta_n^k$ is primitive if and only if $k$ and $n$ are coprime.

The key fact is that $\zeta_n$ is a primitive $n^\text{th}$ root of unity, since its first $n$ powers are distinct. A standard order argument shows that $\zeta_n^a = 1$ if and only if $a|n,$ since writing $n = aq+r,$ $0 \le r < n,$ gives $\zeta_n^r = \zeta_n^{n-aq} = 1,$ but this is impossible unless $r=0.$

If $\text{gcd}(k,n) =d,$ then it is easy to check that $(\zeta_n^k)^{n/d} = 1,$ so if $d>1$ then $\zeta_n^k$ is not primitive. $\big($In fact, it is not hard to show that $\zeta_n^k$ is a primitive $\big(\frac nd\big)^\text{th}$ root of unity.$\big)$

If $\text{gcd}(k,n)=1,$ and $(\zeta_n^k)^r=1,$ then $n|kr,$ but $n$ and $k$ are coprime so $n|r.$ This shows that $\zeta_n^k$ is a primitive $n^\text{th}$ root of unity, because the first positive power of $\zeta_n^k$ that equals $1$ is $(\zeta_n^k)^n.$

Euler's totient function counts the number of positive integers $k \le n$ that are coprime to $n,$ which are precisely the exponents that produce primitive $n^\text{th}$ roots of unity, so this is the number of primitive $n$th roots of unity. $_\square$

Classify the $12^\text{th}$ roots of unity by their multiplicative order.

Let $\zeta_{12} = e^{2\pi i/12}.$ Then the powers of $\zeta_{12}$ are classified according to the GCD of the exponent and $12:$ as in the above proof, $\zeta_{12}^k$ is a $\left(\frac{12}{\gcd (12,k)}\right)^\text{th}$ root of unity.

$\zeta_{12}, \zeta_{12}^5, \zeta_{12}^7, \zeta_{12}^{11}$ are primitive $12^\text{th}$ roots of unity. $($Note $\phi(12)=4.)$

$\zeta_{12}^2, \zeta_{12}^{10}$ are primitive $6^\text{th}$ roots of unity.

$\zeta_{12}^3, \zeta_{12}^9$ are primitive $4^\text{th}$ roots of unity.

$\zeta_{12}^4, \zeta_{12}^8$ are primitive $3^\text{rd}$ roots of unity.

$\zeta_{12}^6 = -1$ is a primitive $2^\text{nd}$ root of unity.

$\zeta_{12}^{12} = 1$ is a primitive $1^\text{st}$ root of unity.

Note that the orders are the divisors of $12.$ There are $\phi(d)$ primitive roots of order $d,$ for each $d|12.$ Since $\sum\limits_{d|12} \phi(d) = 12,$ this accounts for all of the $12^\text{th}$ roots of unity. $_\square$

The above arguments show that the powers of a primitive $n^\text{th}$ root of unity enumerate all the primitive $d^\text{th}$ roots of unity, for all the divisors $d$ of $n.$

Let $\zeta_m$ be a primitive $m^\text{th}$ root of unity, and

let $\zeta_n$ be a primitive $n^\text{th}$ root of unity.

Then $\zeta_m\zeta_n$ is a primitive $\ell^\text{th}$ root of unity for some positive integer $\ell.$

What can we say about $\ell$ in general?

$$

**Clarification:** In the answer choices, $\gcd(\cdot)$ and $\text{lcm}(\cdot)$ denote the greatest common divisor function and the lowest common multiple function, respectively.

## Sum and Product

The product of the primitive $n^\text{th}$ roots of unity is $1$ unless $n=2.$ This is because the set of primitive $n^\text{th}$ roots of unity, $n\ge 3,$ can be partitioned into pairs $\big\{ \zeta^k, \zeta^{n-k} \big\},$ which multiply to give $1.$ $($For $n=2$ this fails because $\zeta^1$ and $\zeta^{2-1}$ coincide.$)$

The sum of the primitive $n^\text{th}$ roots of unity is $\mu(n),$ where $\mu$ is the Möbius function; see that wiki for a proof.

In fact, there is an elegant formula for the sum of the $k^\text{th}$ powers of the primitive $n^\text{th}$ roots of unity:

The sum of the $k^\text{th}$ powers of the primitive $n^\text{th}$ roots of unity is

$\mu(r) \frac{\phi(n)}{\phi(r)},$

where $r = \frac{n}{\gcd (n,k)}.$

The extremal examples of the theorem are when $\text{gcd}(n,k) =1$ and when $\text{gcd}(n,k)=n.$

When $\text{gcd}(n,k)=1,$ taking $k^\text{th}$ powers permutes the primitive $n^\text{th}$ roots of unity, so the sum should still be $\mu(n).$ Indeed, $r=n,$ so $\mu(r)\frac{\phi(n)}{\phi(r)} = \mu(n).$

When $\text{gcd}(n,k)=n,$ the powers are all $1,$ so the sum is $\phi(n).$ In this case $r=1,$ so $\mu(r)\frac{\phi(n)}{\phi(r)} = \phi(n)$ as desired.

Here is an outline: the sum $\sum \zeta^k$ is a sum of primitive $r^\text{th}$ roots of unity, and it runs over all of them. But there are repetitions: the sum has $\phi(n)$ terms and there are $\phi(r)$ primitive $r^\text{th}$ roots of unity, so they are each counted $\frac{\phi(n)}{\phi(r)}$ times. The sum of the primitive $r^\text{th}$ roots of unity is $\mu(r),$ so the result follows. $_\square$

## Group Structure

The $n^\text{th}$ roots of unity form a cyclic group under multiplication, generated by $\zeta_n = e^{2\pi i/n}.$ This group is isomorphic to the additive group ${\mathbb Z}/n$ of the integers mod $n,$ under the isomorphism $\zeta_n^k \mapsto k \pmod n.$ In this context, the primitive $n^\text{th}$ roots of unity correspond via this isomorphism to the other generators of the group. This is the set of multiplicative units $({\mathbb Z}/n)^*$ $($which has $\phi(n)$ elements$).$

The $12^\text{th}$ roots of unity are isomorphic to ${\mathbb Z}/12.$ The primitive $12^\text{th}$ roots of unity, $\zeta_{12}, \zeta_{12}^5, \zeta_{12}^7,$ and $\zeta_{12}^{11},$ correspond to the elements $1,5,7,11$ in ${\mathbb Z}/12,$ respectively. These are the elements which generate ${\mathbb Z}/12$ additively. E.g. the multiples of $5$ are

$5,10,3,8,1,6,11,4,9,2,7,0, \, 5,10,\ldots.$

Every element in ${\mathbb Z}/12$ is on that list.

## Cyclotomic Polynomials

The polynomial $\prod_{\zeta \text{ a primitive } n\text{th root of unity}} (x-\zeta)$ is a polynomial in $x$ known as the $n$th cyclotomic polynomial. It is of great interest in algebraic number theory. For more details and properties, see the wiki on cyclotomic polynomials.

**Cite as:**Primitive Roots of Unity.

*Brilliant.org*. Retrieved from https://brilliant.org/wiki/primitive-roots-of-unity/